Question: What is the extraneous solution to these equations? $\dfrac{x^2 - x}{x + 6} = \dfrac{5x - 9}{x + 6}$
Multiply both sides by $x + 6$ $ \dfrac{x^2 - x}{x + 6} (x + 6) = \dfrac{5x - 9}{x + 6} (x + 6)$ $ x^2 - x = 5x - 9$ Subtract $5x - 9$ from both sides: $ x^2 - x - (5x - 9) = 5x - 9 - (5x - 9)$ $ x^2 - x - 5x + 9 = 0$ $ x^2 - 6x + 9 = 0$ Factor the expression: $ (x - 3)(x - 3) = 0$ Therefore $x = 3$ The original expression is defined at $x = 3$ and $x = 3$, so there are no extraneous solutions.